-0.000 282 006 058 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 058(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 282 006 058(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 282 006 058| = 0.000 282 006 058


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 282 006 058.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 282 006 058 × 2 = 0 + 0.000 564 012 116;
  • 2) 0.000 564 012 116 × 2 = 0 + 0.001 128 024 232;
  • 3) 0.001 128 024 232 × 2 = 0 + 0.002 256 048 464;
  • 4) 0.002 256 048 464 × 2 = 0 + 0.004 512 096 928;
  • 5) 0.004 512 096 928 × 2 = 0 + 0.009 024 193 856;
  • 6) 0.009 024 193 856 × 2 = 0 + 0.018 048 387 712;
  • 7) 0.018 048 387 712 × 2 = 0 + 0.036 096 775 424;
  • 8) 0.036 096 775 424 × 2 = 0 + 0.072 193 550 848;
  • 9) 0.072 193 550 848 × 2 = 0 + 0.144 387 101 696;
  • 10) 0.144 387 101 696 × 2 = 0 + 0.288 774 203 392;
  • 11) 0.288 774 203 392 × 2 = 0 + 0.577 548 406 784;
  • 12) 0.577 548 406 784 × 2 = 1 + 0.155 096 813 568;
  • 13) 0.155 096 813 568 × 2 = 0 + 0.310 193 627 136;
  • 14) 0.310 193 627 136 × 2 = 0 + 0.620 387 254 272;
  • 15) 0.620 387 254 272 × 2 = 1 + 0.240 774 508 544;
  • 16) 0.240 774 508 544 × 2 = 0 + 0.481 549 017 088;
  • 17) 0.481 549 017 088 × 2 = 0 + 0.963 098 034 176;
  • 18) 0.963 098 034 176 × 2 = 1 + 0.926 196 068 352;
  • 19) 0.926 196 068 352 × 2 = 1 + 0.852 392 136 704;
  • 20) 0.852 392 136 704 × 2 = 1 + 0.704 784 273 408;
  • 21) 0.704 784 273 408 × 2 = 1 + 0.409 568 546 816;
  • 22) 0.409 568 546 816 × 2 = 0 + 0.819 137 093 632;
  • 23) 0.819 137 093 632 × 2 = 1 + 0.638 274 187 264;
  • 24) 0.638 274 187 264 × 2 = 1 + 0.276 548 374 528;
  • 25) 0.276 548 374 528 × 2 = 0 + 0.553 096 749 056;
  • 26) 0.553 096 749 056 × 2 = 1 + 0.106 193 498 112;
  • 27) 0.106 193 498 112 × 2 = 0 + 0.212 386 996 224;
  • 28) 0.212 386 996 224 × 2 = 0 + 0.424 773 992 448;
  • 29) 0.424 773 992 448 × 2 = 0 + 0.849 547 984 896;
  • 30) 0.849 547 984 896 × 2 = 1 + 0.699 095 969 792;
  • 31) 0.699 095 969 792 × 2 = 1 + 0.398 191 939 584;
  • 32) 0.398 191 939 584 × 2 = 0 + 0.796 383 879 168;
  • 33) 0.796 383 879 168 × 2 = 1 + 0.592 767 758 336;
  • 34) 0.592 767 758 336 × 2 = 1 + 0.185 535 516 672;
  • 35) 0.185 535 516 672 × 2 = 0 + 0.371 071 033 344;
  • 36) 0.371 071 033 344 × 2 = 0 + 0.742 142 066 688;
  • 37) 0.742 142 066 688 × 2 = 1 + 0.484 284 133 376;
  • 38) 0.484 284 133 376 × 2 = 0 + 0.968 568 266 752;
  • 39) 0.968 568 266 752 × 2 = 1 + 0.937 136 533 504;
  • 40) 0.937 136 533 504 × 2 = 1 + 0.874 273 067 008;
  • 41) 0.874 273 067 008 × 2 = 1 + 0.748 546 134 016;
  • 42) 0.748 546 134 016 × 2 = 1 + 0.497 092 268 032;
  • 43) 0.497 092 268 032 × 2 = 0 + 0.994 184 536 064;
  • 44) 0.994 184 536 064 × 2 = 1 + 0.988 369 072 128;
  • 45) 0.988 369 072 128 × 2 = 1 + 0.976 738 144 256;
  • 46) 0.976 738 144 256 × 2 = 1 + 0.953 476 288 512;
  • 47) 0.953 476 288 512 × 2 = 1 + 0.906 952 577 024;
  • 48) 0.906 952 577 024 × 2 = 1 + 0.813 905 154 048;
  • 49) 0.813 905 154 048 × 2 = 1 + 0.627 810 308 096;
  • 50) 0.627 810 308 096 × 2 = 1 + 0.255 620 616 192;
  • 51) 0.255 620 616 192 × 2 = 0 + 0.511 241 232 384;
  • 52) 0.511 241 232 384 × 2 = 1 + 0.022 482 464 768;
  • 53) 0.022 482 464 768 × 2 = 0 + 0.044 964 929 536;
  • 54) 0.044 964 929 536 × 2 = 0 + 0.089 929 859 072;
  • 55) 0.089 929 859 072 × 2 = 0 + 0.179 859 718 144;
  • 56) 0.179 859 718 144 × 2 = 0 + 0.359 719 436 288;
  • 57) 0.359 719 436 288 × 2 = 0 + 0.719 438 872 576;
  • 58) 0.719 438 872 576 × 2 = 1 + 0.438 877 745 152;
  • 59) 0.438 877 745 152 × 2 = 0 + 0.877 755 490 304;
  • 60) 0.877 755 490 304 × 2 = 1 + 0.755 510 980 608;
  • 61) 0.755 510 980 608 × 2 = 1 + 0.511 021 961 216;
  • 62) 0.511 021 961 216 × 2 = 1 + 0.022 043 922 432;
  • 63) 0.022 043 922 432 × 2 = 0 + 0.044 087 844 864;
  • 64) 0.044 087 844 864 × 2 = 0 + 0.088 175 689 728;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 282 006 058(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1100 1011 1101 1111 1101 0000 0101 1100(2)

6. Positive number before normalization:

0.000 282 006 058(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1100 1011 1101 1111 1101 0000 0101 1100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 006 058(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1100 1011 1101 1111 1101 0000 0101 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1100 1011 1101 1111 1101 0000 0101 1100(2) × 20 =

1.0010 0111 1011 0100 0110 1100 1011 1101 1111 1101 0000 0101 1100(2) × 2-12


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -12

Mantissa (not normalized):
1.0010 0111 1011 0100 0110 1100 1011 1101 1111 1101 0000 0101 1100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1011(10) =

011 1111 0011(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0010 0111 1011 0100 0110 1100 1011 1101 1111 1101 0000 0101 1100 =

0010 0111 1011 0100 0110 1100 1011 1101 1111 1101 0000 0101 1100


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0110 1100 1011 1101 1111 1101 0000 0101 1100


Decimal number -0.000 282 006 058 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 1100 1011 1101 1111 1101 0000 0101 1100

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100